
The perpendicular distance of centroid from any side, 231m
Magnetic field due to one side.
B1=4πrμoI[cosθ1+cosθ2]
=4πr(231)μ0(10)[cos30∘+cos30∘]
=π53μ0[23+23]=π15μ0
Net magnetic field B=3B1=π45μ0=π45×4π×10−7
=180×10−7T
=18μT
JEE Main 2019 — Physics Electromagnetism
The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1m which is carrying a current of 10A is:
[Take μ0=4π×10−7NA−2 ]
Held on 10 Apr 2019 · Verified 6 Jul 2026.
3μT
1μT
18μT
9μT
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